On 09/08/2017 12:00, Laszlo Ersek wrote:
> On 08/09/17 09:26, Paolo Bonzini wrote:
>> On 09/08/2017 03:06, Laszlo Ersek wrote:
>>>>   20.14%  qemu-system-x86_64                  [.] render_memory_region
>>>>   17.14%  qemu-system-x86_64                  [.] subpage_register
>>>>   10.31%  qemu-system-x86_64                  [.] int128_add
>>>>    7.86%  qemu-system-x86_64                  [.] addrrange_end
>>>>    7.30%  qemu-system-x86_64                  [.] int128_ge
>>>>    4.89%  qemu-system-x86_64                  [.] int128_nz
>>>>    3.94%  qemu-system-x86_64                  [.] phys_page_compact
>>>>    2.73%  qemu-system-x86_64                  [.] phys_map_node_alloc
>> Yes, this is the O(n^3) thing.  An optimized build should be faster
>> because int128 operations will be inlined and become much more efficient.
>>> With this patch, I only tested the "93 devices" case, as the slowdown
>>> became visible to the naked eye from the trace messages, as the firmware
>>> enabled more and more BARs / command registers (and inversely, the
>>> speedup was perceivable when the firmware disabled more and more BARs /
>>> command registers).
>> This is an interesting observation, and it's expected.  Looking at the
>> O(n^3) complexity more in detail you have N operations, where the "i"th
>> operates on "i" DMA address spaces, all of which have at least "i"
>> memory regions (at least 1 BAR per device).
> - Can you please give me a pointer to the code where the "i"th operation
> works on "i" DMA address spaces? (Not that I dream about patching *that*
> code, wherever it may live :) )

It's all driven by actions of the guest.

Simply, by the time you get to the "i"th command register, you have
enabled bus-master DMA on "i" devices (so that "i" DMA address spaces
are non-empty) and you have enabled BARs on "i" devices (so that their
BARs are included in the address spaces).

> - You mentioned that changing this is on the ToDo list. I couldn't find
> it under <https://wiki.qemu.org/index.php/ToDo>. Is it tracked somewhere
> else?

I've added it to https://wiki.qemu.org/index.php/ToDo/MemoryAPI (thanks
for the nudge).


> (I'm not trying to urge any changes in the area, I'd just like to learn
> about the code & the tracker item, if there's one.)
> Thanks!
> Laszlo
>> So the total cost is sum i=1..N i^2 = N(N+1)(2N+1)/6 = O(n^3).
>> Expressing it as a sum shows why it gets slower as time progresses.
>> The solution is to note that those "i" address spaces are actually all
>> the same, so we can get it down to sum i=1..N i = N(N+1)/2 = O(n^2).
>> Thanks,
>> Paolo

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